1. Prepared by Lloyd R. Jaisingh
A PowerPoint Presentation Package to Accompany
Applied Statistics in Business & Economics, 4th edition David P. Doane and Lori E. Seward
Prepared by Lloyd R. Jaisingh

2. Probability Chapter Contents Chapter 5 5.1 Random Experiments
5.3 Rules of Probability
5.4 Independent Events
5.5 Contingency Tables
5.6 Tree Diagrams
5.7 Bayes’ Theorem
5.8 Counting Rules

3. Probability Chapter Learning Objectives Chapter 5
LO5-1: Describe the sample space of a random variable.
LO5-2: Distinguish among the three views of probability.
LO5-3: Apply the definitions and rules of probability.
LO5-4: Calculate odds from given probabilities.
LO5-5: Determine when events are independent.

4. Probability Chapter Learning Objectives Chapter 5
LO5-6: Apply the concepts of probability to contingency tables.
LO5-7: Interpret a tree diagram.
LO5-8: Use Bayes’ Theorem to calculate revised probabilities.
LO5-9: Apply counting rules to calculate possible event arrangements.

5. 5.1 Random Experiments Chapter 5 LO5-1
LO5-1: Describe the sample space of a random experiment.
Sample Space
A random experiment is an observational process whose results cannot be known in advance.
The set of all outcomes (S) is the sample space for the experiment.
A sample space with a countable number of outcomes is discrete.

6. 5.1 Random Experiments Sample Space Chapter 5 LO5-1
For a single roll of a die, the sample space is:
When two dice are rolled, the sample space is the following pairs:
5A-6

7. 5.1 Random Experiments Sample Space Chapter 5 LO5-1
If the outcome is a continuous measurement, the sample space cannot be listed but can be described by a rule.
For example, the sample space for the length of a randomly chosen cell phone call would be
S = {all X such that X > 0}.
The sample space to describe a randomly chosen student’s GPA would be
S = {all X such that 0.00 ≤ X ≤ 4.00}.

8. 5.1 Random Experiments Events Chapter 5 LO5-1
An event is any subset of outcomes in the sample space.
A simple event or elementary event, is a single outcome.
A discrete sample space S consists of all the simple events (Ei):
S = {E1, E2, …, En}.
For example, Amazon’s website for “Books & Music”
has seven categories that a shopper might choose:
S = {Books, DVD, VHS, Magazines, Newspapers,
Music, Textbooks}.

9. 5.1 Random Experiments Events Chapter 5 LO5-1
Within this sample space, we could define compound events “electronic media” as A = {Music, DVD, VHS} and “print periodicals” as B = (Newspapers, Magazines}. This can be shown in a Venn diagram.

10. 5.2 Probability Chapter 5 LO5-2
LO5-2: Distinguish among the three views of probability.
Definitions
The probability of an event is a number that measures the relative likelihood that the event will occur.
The probability of event A [denoted P(A)] must lie within the interval from 0 to 1:
0 ≤ P(A) ≤ 1
If P(A) = 0, then the event cannot occur.
If P(A) = 1, then the event is certain to occur.

11. 5.2 Probability Chapter 5 LO5-2 Definitions
In a discrete sample space, the probabilities of all simple events must sum to one:
P(S) = P(E1) + P(E2) + … + P(En) = 1
For example, if on a shopping spree, the following percentages (on the left) were recorded for the four methods of payments. We can compute the equivalent probabilities which sum to 1.
credit card:
32%
debit card:
15%
cash:
35%
check:
18%
Sum =
100%
P(credit card) =
.32
P(debit card) =
.15
P(cash) =
.35
P(check) =
.18
Sum =
1.0
Probability

12. 5.2 Probability Empirical Approach Chapter 5 LO5-2
Use the empirical or relative frequency approach to assign probabilities by counting the frequency (fi) of observed outcomes defined on the experimental sample space.
For example, to estimate the default rate on student loans:
number of defaults
number of loans =
P(a student defaults) = f /n

13. 5.2 Probability Chapter 5 LO5-2 Law of Large Numbers
The law of large numbers says that as the number of trials increases, any empirical probability approaches its theoretical limit.
Flip a coin 50 times. We would expect the proportion of heads to be near .50.
However, in a small finite sample, any ratio can be obtained (e.g., 1/3, 7/13, 10/22, 28/50, etc.).
A large n may be needed to get close to .50.

14. 5.2 Probability
LO5-2
Chapter 5
Law of Large Numbers

15. 5.2 Probability Classical Approach Chapter 5 LO5-2
A priori refers to the process of assigning probabilities before the event is observed or the experiment is conducted.
A priori probabilities are based on logic, not experience.
When flipping a coin or rolling a pair of dice, we do not actually have to perform an experiment because the nature of the process allows us to envision the entire sample space.
Instead of performing the experiment, we can use deduction to determine the probability of an event.
This is the classical approach to probability.

16. 5.2 Probability Classical Approach Chapter 5 LO5-2
For example, the two-dice experiment has 36 equally likely simple events. The P(that the sum of the dots on the two faces equals 7) is
The probability is obtained a priori using the classical approach as shown in this Venn diagram for 2 dice:

17. 5.2 Probability Subjective Approach Chapter 5 LO5-2
A subjective probability reflects someone’s informed judgment about the likelihood of an event.
Used when there is no repeatable random experiment.
For example, - What is the probability that a new truck product program will show a return on investment of at least 10 percent? - What is the probability that the price of Ford’s stock will rise within the next 30 days?

18. 5.3 Rules of Probability Complement of an Event Chapter 5 LO5-3
LO5-3: Apply the definitions and rules of probability.
Complement of an Event
The complement of an event A is denoted by A′ and consists of everything in the sample space S except event A.
Since A and A′ together comprise the entire sample space, P(A) + P(A′ ) = 1 or P(A′ ) = 1 – P(A)

19. 5.3 Rules of Probability Union of Two Events Chapter 5 LO5-3
(Figure 5.5)
The union of two events consists of all outcomes in the sample space S that are contained either in event A or in event B or in both (denoted A  B or “A or B”).
 may be read as “or” since one or the other or both events may occur.

20. 5.3 Rules of Probability Intersection of Two Events Chapter 5 LO5-3
The intersection of two events A and B (denoted by A  B or “A and B”) is the event consisting of all outcomes in the sample space S that are contained in both event A and event B.
 may be read as “and” since both events occur. This is a joint probability.

21. 5.3 Rules of Probability General Law of Addition Chapter 5 LO5-3
The general law of addition states that the probability of the union of two events A and B is:
P(A  B) = P(A) + P(B) – P(A  B)
When you add the P(A) and P(B) together, you count the P(A and B) twice.
A and B
So, you have to subtract P(A  B) to avoid overstating the probability. A B

22. 5.3 Rules of Probability General Law of Addition Chapter 5 LO5-3
For a standard deck of cards:
P(Q) = 4/52 (4 queens in a deck; Q = queen)
P(R) = 26/52 (26 red cards in a deck; R = red)
P(Q  R) = 2/52 (2 red queens in a deck)
P(Q  R) = P(Q) + P(R) – P(Q  R)
Q and R = 2/52
= 4/ /52 – 2/52
Q 4/52
R 26/52
= 28/52 = or 53.85%

23. 5.3 Rules of Probability Mutually Exclusive Events Chapter 5 LO5-3
Events A and B are mutually exclusive (or disjoint) if their intersection is the null set () which contains no elements.
If A  B = , then P(A  B) = 0
Special Law of Addition
In the case of mutually exclusive events, the addition law reduces to:
P(A  B) = P(A) + P(B)

24. 5.3 Rules of Probability Collectively Exhaustive Events Chapter 5
LO5-3
Chapter 5
Collectively Exhaustive Events
Events are collectively exhaustive if their union is the entire sample space S.
Two mutually exclusive, collectively exhaustive events are dichotomous (or binary) events.
For example, a car repair is either covered by the warranty (A) or not (A’).
Note: This concept can be extended to more than two events. See the next slide
No Warranty
Warranty

25. 5.3 Rules of Probability Collectively Exhaustive Events Chapter 5
LO5-3
Chapter 5
Collectively Exhaustive Events
There can be more than two mutually exclusive, collectively exhaustive events, as illustrated below. For example, a Walmart customer can pay by credit card (A), debit card (B), cash (C), or
check (D).

26. 5.3 Rules of Probability Conditional Probability Chapter 5 LO5-3
The probability of event A given that event B has occurred.
Denoted P(A | B). The vertical line “ | ” is read as “given.”

27. 5.3 Rules of Probability Conditional Probability Chapter 5 LO5-3
Consider the logic of this formula by looking at the Venn diagram.
The sample space is restricted to B, an event that has occurred.
A  B is the part of B that is also in A.
The ratio of the relative size of A  B to B is P(A | B).

28. 5.3 Rules of Probability Example: High School Dropouts Chapter 5 LO5-3
Of the population aged 16–21 and not in college:
Unemployed
13.5%
High school dropouts
29.05%
Unemployed high school dropouts
5.32%
What is the conditional probability that a member of this population is unemployed, given that the person is a high school dropout?

29. 5.3 Rules of Probability Example: High School Dropouts Chapter 5 LO5-3
First define
U = the event that the person is unemployed
D = the event that the person is a high school dropout
P(U) = .1350
P(D) = .2905
P(UD) = .0532
P(U | D) = > P(U) = .1350
Therefore, being a high school dropout is related to being unemployed.

30. 5.3 Rules of Probability Chapter 5 LO5-4
LO5-4: Calculate odds from given probabilities
Odds of an Event
The odds in favor of event A occurring are
The odds against event A occurring are
5A-30 30

31. 5.3 Rules of Probability Odds of an Event Chapter 5 LO5-4
If the odds against event A are quoted as b to a, then the implied probability of event A is:
For example, if a race horse has a 4 to 1 odds against winning, the P(win) is
5A-31

32. 5.4 Independent Events Chapter 5 LO5-5
LO5-5: Determine when events are independent
Event A is independent of event B if the conditional probability P(A | B) is the same as the marginal probability P(A).
P(U | D) = > P(U) = .1350, so U and D are not independent. That is, they are dependent.
Another way to check for independence: Multiplication Law
If P(A  B) = P(A)P(B) then event A is independent of event B since

33. 5.4 Independent Events
LO5-5
Chapter 5
Application of the Multiplication Law (for Independent Events)
The probability of n independent events occurring simultaneously is:
P(A1  A2  ...  An) = P(A1) P(A2) ... P(An)
if the events are independent
To illustrate system reliability, suppose a website has 2 independent file servers. Each server has 99% reliability. What is the total system reliability? Let
F1 be the event that server 1 fails
F2 be the event that server 2 fails

34. 5.4 Independent Events
LO5-5
Chapter 5
Application of the Multiplication Law (for Independent Events)
Applying the rule of independence:
P(F1  F2 ) = P(F1) P(F2) = (.01)(.01) = .0001
So, the probability that both servers are down is
The probability that one or both servers is “up” is:
= or 99.99%

35. 5.5 Contingency Table Chapter 5 LO5-6
LO5-6: Apply the concepts of probability to contingency tables.
Example: Salary Gains and MBA Tuition
Consider the following cross-tabulation (contingency) table for n = 67 top-tier MBA programs: 35

36. 5.5 Contingency Table Example: Salary Gains and MBA Tuition Chapter 5
LO5-6
Chapter 5
Example: Salary Gains and MBA Tuition
Are large salary gains more likely to accrue to graduates of high-tuition MBA programs?
The frequencies indicate that MBA graduates of high-tuition
schools do tend to have large salary gains.
Also, most of the top-tier schools charge high tuition.
More precise interpretations of these data can be made using the concepts of probability.

37. 5.5 Contingency Table Marginal Probabilities Chapter 5 LO5-6
The marginal probability of a single event is found by dividing a row or column total by the total sample size.
For example, find the marginal
probability of a medium salary gain (P(S2).
P(S2) = 33/67 = .4925
Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).

38. 5.5 Contingency Table Chapter 5 LO5-6 Marginal Probabilities
Find the marginal probability of a medium salary gain P(S2).
P(S2) = 33/67 = .4925
There is a 49% chance that a top-tier school’s MBA student will experience a medium salary gain. 38

39. 5.5 Contingency Table Joint Probabilities Chapter 5 LO5-6
A joint probability represents the intersection of two events in a cross-tabulation table.
Consider the joint event that the school has low tuition and large salary gains (denoted as P(T1  S3)).
P(T1  S3) = 1/67 = .0149
There is less than a 2% chance that a top-tier school has both low tuition and large salary gains. 39

40. 5.5 Contingency Table Conditional Probabilities Independence Chapter 5
LO5-6
Chapter 5
Conditional Probabilities
Find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3).
P(T3 | S1) = 5/32 = .1563
Independence
Conditional
Marginal
P(S3 | T1)= 1/16 = .0625
P(S3) = 17/67 = .2537
(S3) and (T1) are dependent. 40

41. 5.5 Contingency Table Relative Frequencies Chapter 5 LO5-6
Here are the resulting probabilities (relative frequencies). Each value is divided by 67. For example,
P(T1 and S1) = 5/67
P(T2 and S2) = 11/67
P(T3 and S3) = 15/67
P(S1) = 17/67
P(T2) = 19/67 41

42. 5.6 Tree Diagrams What is a Tree? Chapter 5 LO5-7
LO5-7: Interpret a tree diagram.
What is a Tree?
A tree diagram or decision tree helps you visualize all possible outcomes.
Start with a contingency table.
For example, this table gives expense ratios by fund type for 21 bond funds and 23 stock funds.
The tree diagram shows all events along with their marginal, conditional, and joint probabilities.

43. 5.6 Tree Diagrams Chapter 5 LO5-7
Tree Diagram for Fund Type and Expense Ratios

44. 5.7 Bayes’ Theorem Chapter 5 LO5-8
LO8: Use Bayes’ Theorem to compute revised probabilities
Thomas Bayes ( ) provided a method (called Bayes Theorem) of revising probabilities to reflect new probabilities.
The prior (marginal) probability of an event B is revised after event A has been considered to yield a posterior (conditional) probability.
Bayes’s formula is:
In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is:

45. 5.7 Bayes’ Theorem General Form of Bayes’ Theorem Chapter 5 LO5-8
A generalization of Bayes’s Theorem allows event B to have as many
mutually exclusive and collectively exhaustive categories as we wish
(B1, B2, …, Bn) rather than just two dichotomous categories (B and B').

46. 5.7 Bayes’ Theorem Chapter 5 LO5-8 Example: Hospital Trauma Centers
(Table 5.18)
Based on historical data, the percent of cases at 3 hospital trauma centers and the probability of a case resulting in a malpractice suit are as follows:
Let event A = a malpractice suit is filed Bi = patient was treated at trauma center i

47. 5.7 Bayes’ Theorem Example: Hospital Trauma Centers Chapter 5 LO5-8
Applying the general form of Bayes’ Theorem, find P(B1 | A).

48. 5.7 Bayes’ Theorem Chapter 5 LO5-8 Example: Hospital Trauma Centers
Conclude that the probability that the malpractice suit was filed in hospital 1 is .1389, or 13.89%.
All the posterior probabilities for each hospital can be calculated and then compared:

49. 5.8 Counting Rules Fundamental Rule of Counting
LO5-9
Chapter 5
LO9: Apply counting rules to calculate possible event arrangements.
Fundamental Rule of Counting
If event A can occur in n1 ways and event B can occur in n2 ways, then events A and B can occur in n1 x n2 ways.
In general, m events can occur n1 x n2 x … x nm ways.
Example: Stockkeeping Labels
How many unique stockkeeping unit (SKU) labels can a hardware store create by using two letters (ranging from AA to ZZ) followed by four numbers (0 through 9)? 49

50. 5.8 Counting Rules Chapter 5 LO5-9 Example: Stockkeeping Labels
For example, AF1078: hex-head 6 cm bolts – box of 12; RT4855: Lime-A-Way cleaner – 16 ounce LL3319: Rust-Oleum primer – gray 15 ounce
There are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels. 50

51. 5.8 Counting Rules Factorials Chapter 5 LO5-9
The number of ways that n items can be arranged in a particular order is n factorial.
n factorial is the product of all integers from 1 to n.
n! = n(n–1)(n–2)...1
Factorials are useful for counting the possible arrangements of any n items.
There are n ways to choose the first, n-1 ways to choose the second, and so on.
A home appliance service truck must make 3 stops (A, B, C). In how many ways could the three stops be arranged?
3! = 3 x 2 x 1 = 6 51

52. 5.8 Counting Rules Permutations Chapter 5 LO5-9
A permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by nPr
In other words, how many ways can the r items be arranged from n items, treating each arrangement as different (i.e., XYZ is different from ZYX)? 52

53. 5.8 Counting Rules Chapter 5 LO5-9 Combinations
A combination is an arrangement of r items chosen at random from n items where the order of the selected items is not important (i.e., XYZ is the same as ZYX).
A combination is denoted nCr 53